Find the series expansion of \(\frac{1}{\sqrt{1+5x}}\), up to and including the term in \(x^2\). State the range of values of \(x\) for which the expansion is valid.[3]
Hence, use the substitution \(x = \frac{1}{20}\) to obtain an approximation of \(\sqrt{5}\), expressing your answer as a fraction.[2]
Without any further calculation, explain whether using the substitution \(x = -\frac{4}{25}\) gives a better approximation of \(\sqrt{5}\) than the substitution used in part (b).[1]
Solution:
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Answer:(a) \(1-\frac52x+\frac{75}{8}x^2+\cdots\), \(-\frac15<x<\frac15\). (b) \(\sqrt5\approx\frac{575}{256}\); the reciprocal route gives \(\frac{256}{115}\). (c) No; the substitution is farther from zero.